tìm x,biết
a, 2x.4=128
b, (2x+1)=125
c, x15=x
d
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a) \(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
c) \(\Rightarrow2^x=32\Rightarrow x=5\)
d) \(\Rightarrow4^3.4^x=4^5\Rightarrow4^x=4^2\Rightarrow x=2\)
e) \(\Rightarrow3^3.3^x=3^5\Rightarrow3^x=3^2\Rightarrow x=2\)
f) \(\Rightarrow7^2.7^x=7^4\Rightarrow7^x=7^2\Rightarrow x=2\)
a. 2x . 4 = 128
<=> 2x + 2 = 27
<=> x + 2 = 7
<=> x = 5
b. (2x + 1)3 = 125
<=> (2x + 1)3 - 53 = 0
<=> (2x + 1 - 5)\(\left[\left(2x+1\right)^2+\left(2x+1\right).5+25\right]=0\)
<=> (2x - 4)(4x2 + 4x + 1 + 10x + 5 + 25) = 0
<=> (2x - 4)(4x2 + 14x + 31) = 0
<=> \(\left[{}\begin{matrix}2x-4=0\\4x^2+14x+31=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\VôNghiệm\end{matrix}\right.\)
c. 2x - 26 = 6
<=> 2x = 32
<=> x = 5
d. 64 . 4x = 45
<=> 43 . 4x = 45
<=> 43 + x = 45
<=> 3 + x = 5
<=> x = 2
e. 27 . 3x = 243
<=> 33 . 3x = 35
<=> 33 + x = 35
<=> 3 + x = 5
<=> x = 2
g. 49 . 7x = 2401 (Bn xem lại đề câu này)
<=> 72 . 7x = 74
<=> 72 + x = 74
<=> 2 + x = 4
<=> x = 2
a) \(2^x\cdot4=128\)
\(\Rightarrow2^x\cdot2^2=2^7\)
\(\Rightarrow2^{x+2}=2^7\)
\(\Rightarrow x+2=7\)
\(\Rightarrow x=5\)
b) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=4:2\)
\(\Rightarrow x=2\)
c) \(2x-2^6=6\)
\(\Rightarrow2x-64=6\)
\(\Rightarrow2x=70\)
\(\Rightarrow x=70:2\)
\(\Rightarrow x=35\)
d) \(64\cdot4^x=45\)
\(\Rightarrow4^3\cdot4^x=45\)
\(\Rightarrow4^{x+3}=45\)
Xem lại đề
e) \(27\cdot3^x=243\)
\(\Rightarrow3^3\cdot3^x=3^5\)
\(\Rightarrow3^{x+3}=3^5\)
\(\Rightarrow x+3=5\)
\(\Rightarrow x=2\)
g) \(49\cdot7^x=2401\)
\(\Rightarrow7^2\cdot7^x=7^4\)
\(\Rightarrow7^{x+2}=7^4\)
\(\Rightarrow x+2=4\)
\(\Rightarrow x=2\)
h) \(3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
k) \(3^4\cdot3^x=3^7\)
\(\Rightarrow3^{x+4}=3^7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
n) \(3^x+25=26\cdot2^2+2\cdot3^0\)
\(\Rightarrow3^x+25=104+2\)
\(\Rightarrow3^x+25=106\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(x=4\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`2^x*4 = 128`
`=> 2^x = 128 \div 4`
`=> 2^x = 2^7 \div 2^2`
`=> 2^x = 2^5`
`=> x = 5`
Vậy, `x = 5.`
`b)`
\(\left(2x+1\right)^3=125\)
`=> (2x + 1)^3 = 5^3`
`=> 2x + 1 = 5`
`=> 2x = 5-1`
`=> 2x = 4`
`=> x = 4 \div 2`
`=> x = 2`
Vậy, `x = 2`
`c)`
\(2x-2^6=6\)
`=> 2x = 6+2^6`
`=> 2x = 70`
`=> x = 70 \div 2`
`=> x = 35`
Vậy, `x = 35`
`d)`
\(64\cdot4^x=45\) Bạn xem lại đề
`e)`
`27*3^x = 243`
`=> 3^3 * 3^x = 3^5`
`=> 3^(3 + x) = 3^5`
`=> 3 + x = 5`
`=> x = 5 - 3`
`=> x = 2`
Vậy, `x = 2`
`g)`
`49* 7^x = 2401`
`=> 7^2 * 7^x = 7^4`
`=> 7^(2 + x) = 7^4`
`=> 2 + x = 4`
`=> x = 4 - 2`
`=> x = 2`
Vậy, `x = 2`
`h)`
`3^x = 81`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4`
`k)`
`3^4 * 3^x = 3^7`
`=> 3^(4 + x) = 3^7`
`=> 4 + x = 7`
`=> x = 7 - 4`
`=> x = 3`
Vậy, `x = 3`
`n)`
`3^x + 25 = 26*2^2 + 2*3^0`
`=> 3^x + 25 = 104 + 2`
`=> 3^x + 25 = 106`
`=> 3^x = 106 - 25`
`=> 3^x = 81`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4.`
\(#48Cd\)
a: =>12x-64=32
=>12x=96
=>x=8
b: =>x-1=5
=>x=6
c: =>2^x*3=96
=>2^x=32
=>x=5
2:
a: =>2(x+1)=26
=>x+1=13
=>x=12
b: =>(6x)^3=125
=>6x=5
=>x=5/6(loại)
c: =>\(7\cdot3^x\cdot\dfrac{1}{3}+11\cdot3^x\cdot3=318\)
=>3^x=9
=>x=2
d: -2x+13 chia hết cho x+1
=>-2x-2+15 chia hết cho x+1
=>15 chia hết cho x+1
=>x+1 thuộc {1;3;5;15}
=>x thuộc {0;2;4;14}
e: 4x+11 chia hết cho 3x+2
=>12x+33 chia hết cho 3x+2
=>12x+8+25 chia hết cho 3x+2
=>25 chia hết cho 3x+2
=>3x+2 thuộc {1;-1;5;-5;25;-25}
mà x là số tự nhiên
nên x=1
1:
a: Đặt A=2^2024-2^2023-...-2^2-2-1
Đặt B=2^2023+2^2022+...+2^2+2+1
=>2B=2^2024+2^2023+...+2^3+2^2+2
=>B=2^2024-1
=>A=2^2024-2^2024+1=1
c: \(=\dfrac{3^{12}\cdot2^{11}+2^{10}\cdot3^{12}\cdot5}{2^2\cdot3\cdot3^{11}\cdot2^{11}}=\dfrac{2^{10}\cdot3^{12}\left(2+5\right)}{2^{13}\cdot3^{12}}\)
\(=\dfrac{7}{2^3}=\dfrac{7}{8}\)
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
a: -2x(x+3)+x(2x-1)=10
=>-2x^2-6x+2x^2-x=10
=>-7x=10
=>x=-10/7
b: Sửa đề: 2/3x(9/2x+1/4)-(3x^2+2)=3
=>3x^2+1/6x-3x^2-2=3
=>1/6x-2=3
=>x=30
a: =>3^x=3^4*3=3^5
=>x=5
b: =>\(2^{x+1}=2^5\)
=>x+1=5
=>x=4
c: \(\Leftrightarrow3^{x+2-3}=3\)
=>x-1=1
=>x=2
d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)
=>x=4 hoặc x=-4
e: (2x-1)^4=81
=>2x-1=3 hoặc 2x-1=-3
=>2x=4 hoặc 2x=-2
=>x=-1 hoặc x=2
f: (2x-6)^4=0
=>2x-6=0
=>x-3=0
=>x=3
a) \(3^x=81\cdot3\)
\(\Rightarrow3^x=3^4\cdot3\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
b) \(2^{x+1}=32\)
\(\Rightarrow2^{x+1}=2^5\)
\(\Rightarrow x+1=5\)
\(\Rightarrow x=4\)
c) \(3^{x+2}:27=3\)
\(\Rightarrow3^{x+2}:3^3=3\)
\(\Rightarrow3^{x+2-3}=3\)
\(\Rightarrow3^{x-1}=3\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
d) \(2x^2=32\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
e) \(\left(2x-1\right)^4=81\)
\(\Rightarrow\left(2x-1\right)^4=3^4\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
f) \(\left(2x-6\right)^4=0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=6:2\)
\(\Rightarrow x=3\)
a. 2x.4=128
2x =32
=> x = 5
b. 2x+1=125
2x = 125-1
2x = 124
x = 62
c. x15=x
=> x \(\in\left\{0;\pm1\right\}\)